1 g of activated charcoal has a surface area 105 m 2. If there is complete monolayer coverage and has effective surface area of NH 3 is 0.015 nm 2, then volume of NH 3 in cm 3 at STP that could be adsorbed on 50 g of charcoal is .

Question

1 g of activated charcoal has a surface area 105 m 2. If there is complete monolayer coverage and has effective surface area of NH 3 is 0.015 nm 2, then volume of NH 3 in cm 3 at STP that could be adsorbed on 50 g of charcoal is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total surface area available for adsorption

= 50 \times 1 0^{3} m^{2}

Effective surface area of one

N H_{3}

molecule

= 1.5 \times 1 0^{- 20} m^{2}

∴

No. of molecules of

N H_{3}

adsorbed

= \frac{50 \times 1 0 ^{3}}{1.5 \times 1 0 ^{- 20}}

= 33.3 \times 1 0^{17}

Moles of

N H_{3} = \frac{33.3 \times 1 0 ^{17}}{6 \times 1 0 ^{23}}

= 5.5 \times 1 0^{- 6}

∴ V^{N H_{3}} = \frac{n RT}{P}

= \frac{5.5 \times 1 0 ^{- 6} \times 0.082 \times 273}{1}

= 1.23 \times 1 0^{- 4} l i t re

= 0.123 c m^{3}

Q. 1g of activated charcoal has a surface area 105m2. If there is complete monolayer coverage and has effective surface area of NH3​ is 0.015nm2, then volume of NH3​ in cm3 at STP that could be adsorbed on 50g of charcoal is _______.

Q. $1 g$ of activated charcoal has a surface area $1 0^{5} m^{2}$ . If there is complete monolayer coverage and has effective surface area of $N H_{3}$ is $0.015 n m^{2}$ , then volume of $N H_{3}$ in $c m^{3}$ at STP that could be adsorbed on $50 g$ of charcoal is _______.